Lorentz Transformation

Textbook & Collegerama

• Chapter 11: section 11.3 & 11.4
• TN1612TU_25

Learning goals

• Write down the Lorentz transformation
• Understand the starting point of the derivation
• Understand the meaning of the Lorentz factor

Best form of the Lorentz transformation

The way we will write down the Lorentz transformation is a bit particular in a sense that we combine time \(t\) with the speed of light \(c\) into the "time" axis \(ct\) which now has unit length. We can do this as \(c\) is constant for all observers independent of their frame of reference. The speed of light is said to be a Lorentz invariant. In this notation the transform between \(S\) and \(S'\) (moving with velocity \(V\) away) is easy to remember!

\[ \begin{array}{rcl} ct' &=& \gamma\left (ct-\frac{V}{c}x \right ) \\ x' &=& \gamma\left(x-\frac{V}{c}ct \right )\\ y' &=& y\\ z' &=& z \end{array} \]

The inverse is given by

\[ \begin{array}{rcl} ct &=& \gamma(ct'+\frac{V}{c}x') \\ x &=& \gamma(x'+\frac{V}{c}ct')\\ y &=& y'\\ z &=& z' \end{array} \]

with the Lorentz factor \(\gamma(V) \equiv \frac{1}{\sqrt{1-\frac{V^2}{c^2}}} \geq 1\).The structure of the formulas is very symmetric and therefore needs little remembering.

From the Lorentz transformation it is clear that time is not universal anymore (\(ct'\neq ct\) in general). This is a large step from Newton and Galileo. Now the time coordinated is mixed somehow with the space coordinated depending on the speed \(V\).

Lorentz factor

The Lorentz factor (or \(\gamma\)-factor)

\[ \gamma=\frac{1}{\sqrt{1-\frac{V^2}{c^2}}} \geq 1 \]

is a dimensionless constant depending on the ratio of the velocity \(V\) to the speed of light \(c\). Sometimes this ratio \(V/c\) is abbreviated further as \(\beta \equiv \frac{V}{c} \leq 1\). For the ratio we know that it is smaller than 1 as \(c\) is a limit velocity. From that it follows that the \(\gamma\)-factor is always equal or larger to one, \(\gamma \geq 1\). As we will see later the factor also relates the coordinated time interval \(dt\) to the proper time \(d\tau\), as \(\gamma = \frac{dt}{d\tau}\).

In many exercises the speed \(V\) is given already as fraction of \(c\), e.g. \(V=0.8c\). Analytically only for very few speeds a nice \(\gamma\)-factor is computed. These are

\[ \begin{array}{rcl} V = \frac{3}{5}c &\Leftrightarrow & \gamma = \frac{5}{4}\\ V = \frac{4}{5}c &\Leftrightarrow & \gamma = \frac{5}{3}\\ V = \frac{12}{13}c &\Leftrightarrow & \gamma = \frac{13}{5} \end{array} \]

In the limit

In the limit of low speeds with respect to the speed of light \(\frac{V}{c}\ll 1 \Rightarrow \gamma =1\). Practically, this happens for about \(V< 0.1 c \sim 30.000\) km/s. In this limit the Lorentz transformation also reduces to the Galileo transformation.

\[ \begin{array}{rcl} ct' &=& ct \\ x' &=& x-Vt \\ y' &=& y\\ z' &=& z \end{array} \]

In the limit of infinity speed of light (\(c\to\infty\)) the \(\gamma\)-factor is again one \(\gamma=1\) and the ratio \(V/c \to 0\). Also here the LT reduces to the GT. The case of infinite speed of light represents the case that GT is generally valid, i.e. \(c'=c+V\).

It is always important to verify that an extension of the known theory reduces to the known theory that has proofed itself for most circumstances.

Derivation of the Lorentz Transformation

The derivation we give is a bit more detailed in the steps than the one given by Einstein himself in Über die spezielle und allgemeine Relativitätstheorie. The book is written such that it can be followed easily with high-school math but is nevertheless quite some work (as is this derivation).

We start by the constant of the speed of light for all observers.

We have observers \(S\) and \(S'\), where \(S'\) moves away from \(S\) with velocity \(V\). Consider a light wave traveling in the positive \(x\)-direction which is describe in both systems as \(x-ct=0\) or \(x'-ct'=0\). NB: In 3D a light wave is described by \(x^2+y^2+z^2-(ct)^2=0\) which reduced to \(x-ct=0\) in 1D.

Therefore the transformation relating both systems must be of the form (with \(\lambda\) a real constant)

\[ (x'-ct') = \lambda (x-ct)\quad (*) \]

If we consider a light wave traveling in the negative \(x\)-direction then, it is described in both systems as \(x+ct=0\) and \(x'+ct'=0\) and we need a relation of the form (with real constant \(\mu\))

\[ (x'+ct') = \mu (x+ct)\quad (**) \]

Now we add both equations \((*)+(**)\) which gives

\[ \begin{array}{rcl} 2x' &=& \lambda (x-ct)+\mu(x+ct)\\ x' &=& x\left ( \frac{\lambda+\mu}{2}\right ) -ct\left ( \frac{\lambda-\mu}{2}\right ) \end{array} \]

If we subtract the two equations \((*)-(**)\)

\[ \begin{array}{rcl} 2ct' &=& \lambda (x-ct)-\mu(x+ct)\\ ct' &=& ct\left ( \frac{\lambda+\mu}{2}\right ) -x\left ( \frac{\lambda-\mu}{2}\right ) \end{array} \]

For connivance we redefine the two constants here \(a=\frac{\lambda+\mu}{2}, b=\frac{\lambda-\mu}{2}\) and the transformation takes the form

\[ \begin{array}{rcl} ct' &=& act-bx\\ x' &=& ax-bct \end{array} \]

Now we consider that the relative speed is \(V\) and that in the origin of \(S': x'=0\) thus \(x=\frac{bc}{a}t\) from the last equation. The velocity of \(S'\) with respect to \(S\) is thus \(V=\frac{bc}{a}\). We have eliminated the constant \(b=\frac{V}{c}a\) by taking in the velocity.

\[ \begin{array}{rcl} ct' &=& a(ct-\frac{V}{c}x) \quad (\#)\\ x' &=& a(x-\frac{V}{c}ct) \quad (\#\#) \end{array} \]

Now we need only to find the constant \(a\).

1. We therefore consider a rod of length 1 that is at rest in frame \(S'\) as seen from \(S\), thus we have \(\Delta x'=1\) on time point \(t=0\). Using equation \((\#\#)\) we obtain therefore \(\Delta x=\frac{1}{a}\quad (+)\).
2. We also consider a rod of length 1 that is at rest in frame \(S\) as seen from \(S'\). For \(t'=0\) we have \(0=a(ct-\frac{V}{c}x)\) from equation \((\#)\). From equation \((\#\#)\) we have \(act=\frac{c}{V}(ax-x')\). Putting both equations together we have

\[ \begin{array}{rcl} 0 &=& \frac{c}{V}(ax-x')-\frac{V}{c}ax\\ x' &=& a(1-\frac{V^2}{c^2}x) \Rightarrow \Delta x' = a(1-\frac{V^2}{c^2}) \quad (++) \end{array} \]

Now we invoke the relativity principle that the interval given by equation \((+)\) and \((++)\) must be the same for both observers! This means that there is no preferential observer and what holds for one must hold for the other, as by axion 1.

\[ \frac{1}{a} = a \left ( 1-\frac{V^2}{c^2}\right ) \Leftrightarrow a=\frac{1}{\sqrt{1-\frac{V^2}{c^2}}} \equiv \gamma (V) \]

The constant \(a\) has been eliminated and replaced by the ratio of the relative velocity \(V\) and the speed of light \(c\). As this is a complicated factor we write short hand \(\gamma\) for it. Now we have arrived at the Lorentz transformation

\[ \begin{array}{rcl} ct' &=& \gamma(ct-\frac{V}{c}x) \\ x' &=& \gamma(x-\frac{V}{c}ct) \end{array} \]

For the inverse transformation, if you want \(ct,x\) explicitly instead of implicitly the transformation reads

\[ \begin{array}{rcl} ct &=& \gamma(ct'+\frac{V}{c}x') \\ x &=& \gamma(x'+\frac{V}{c}ct') \end{array} \]

The easiest way to see this (without solving the math explicitly), is to think \(ct',x' \to ct,x\) and therefore \(V\to -V\).

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Please notice that somehow it was natural to consider the time axis as \(ct\) (instead of \(t\)) which is of course a length unit not a time unit. Later we see that this allows calculations in spacetime (nearly) with vectors as you are used to as the units of all coordinates are in meters.

Historical context

Lorentz did not derive the transformation that now has his name, based on the same ideas as we did here starting from Einstein's axioms. Lorentz, however, saw that Maxwell's equations were not GT invariant, therefore he tried to find a transformation under which they were invariant. He did so (with a bit of help from Poincaré afterwards). FitzGerald did also derive the transformation, but too did not understand its implications.

Before Einstein's idea spread, Lorentz thought about the transformation as a fix to Galileo Transformation. Later he understood of course.

The wave equation is Lorentz invariant

The Navier-Stokes equation was Galileo invariant as shown earlier, but the Maxwell equations not. Here we will show that the wave equation is Lorentz invariant. It is also clear that the Navier-Stokes equation is then not correct for high speeds as it is not Lorentz invariant.

We remember that we need to transform the second derivatives of \(S\) with respect to space and time to the new coordinates of \(S'\) as the wave equation for the field \(\phi(x,y)\) is \(\frac{\partial^2}{\partial x^2}\phi(x,t) =\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\phi(x,t)\).

The old derivatives expressed in the new coordinates of \(S'\) are

\[ \begin{array}{rcl} \displaystyle{\frac{\partial}{\partial x}} &=& \displaystyle{\frac{\partial}{\partial x'}\frac{\partial x'}{\partial x} + \frac{\partial}{\partial t'}\frac{\partial t'}{\partial x} = \gamma \frac{\partial}{\partial x'} - \gamma \frac{V}{c^2}\frac{\partial}{\partial t'}}\\ \displaystyle{\frac{\partial}{\partial t}} &=& \displaystyle{\frac{\partial}{\partial x'}\frac{\partial x'}{\partial t} + \frac{\partial}{\partial t'}\frac{\partial t'}{\partial t} =\gamma \frac{\partial}{\partial t'} - \gamma V \frac{\partial}{\partial x'}} \end{array} \]

Applying the partial derivative twice we arrive at

\[ \begin{array}{rcl} \displaystyle{\frac{\partial^2}{\partial x^2}} &=& \displaystyle{\gamma^2 \frac{\partial^2}{\partial x'^2} -2\gamma^2 \frac{V}{c^2}\frac{\partial^2}{\partial t'\partial x'}+\gamma^2 \frac{V^2}{c^4}\frac{\partial^2}{\partial t'^2}}\\ \displaystyle{\frac{1}{c^2}\frac{\partial^2}{\partial t^2}} &=& \displaystyle{\gamma^2 \frac{V^2}{c^2} \frac{\partial^2}{\partial x'^2}-2\gamma^2 \frac{V}{c^2} \frac{\partial^2}{\partial t'\partial x'}+\frac{\gamma^2}{c^2} \frac{\partial^2}{\partial t'^2}} \end{array} \]

If we fill in these derivative in the wave equation, we see that the factor \(1/c^2\) on the side of the second time derivative cancels out the mixed term. In addition we see, that each term has \(\gamma^2\) in front which we can cross out. Now we sort all spatial derivatives to one side, and all time derivatives to the other side. We remain with

\[ \begin{array}{rcl} \displaystyle{\frac{\partial^2}{\partial x'^2}\phi(x,t)\left ( 1-\frac{V^2}{c^2} \right )}&=& \displaystyle{\frac{1}{c^2}\frac{\partial^2}{\partial t'^2}\phi(x,t)\left ( 1-\frac{V^2}{c^2}\right )}\\ \displaystyle{\frac{\partial^2}{\partial x'^2} \phi(x,t)}&=& \displaystyle{\frac{1}{c^2}\frac{\partial^2}{\partial t'^2}\phi(x,t)} \end{array} \]

The wave equation is Lorentz invariant, with constant speed of light \(c=c'\). From this derivation we also see that the value of the \(\gamma\) factor is not relevant for the Lorentz invariance as it cancels out in the derivatives.

The Maxwell's equations as a whole are Lorentz invariant, which you can now prove yourself if you have the time.